A parallel plate capacitor is charged by a battery. The battery is removed and a thick glass slab is inserted between the plates. Now,

Options

A. The capacity of the capacitor is increased
B. The electrical energy stored in the capacitor is decreased
C. The potential across the plate is decreased
D. The electric field between the plates is decreased.

Answer

To solve the problem regarding the effects on a parallel plate capacitor when a thick glass slab is inserted after charging and disconnecting from the battery, we will analyze the changes in capacitance, voltage, electric field, and energy stored in the capacitor step by step.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions:**
- A parallel plate capacitor is charged by a battery, resulting in a charge \( +Q \) on one plate and \( -Q \) on the other plate.
- The capacitance \( C_0 \) of the capacitor without the dielectric is given by:
\[ C_0 = \frac{\varepsilon_0 A}{d} \]
where \( \varepsilon_0 \) is the permittivity of free space, \( A \) is the area of the plates, and \( d \) is the separation between the plates.

2. **Removing the Battery:**
- Once the battery is removed, the charge \( Q \) on the capacitor plates remains constant.

3. **Inserting the Glass Slab:**
- A thick glass slab (dielectric) with dielectric constant \( K \) is inserted between the plates. The presence of the dielectric affects the capacitance and electric field.

4. **Effect on Capacitance:**
- The new capacitance \( C \) with the dielectric inserted is given by:
\[ C = K \cdot C_0 = K \cdot \frac{\varepsilon_0 A}{d} \]
- Since \( K > 1 \), the capacitance increases.

5. **Effect on Electric Field:**
- The electric field \( E \) between the plates without the dielectric is given by:
\[ E_0 = \frac{V_0}{d} \]
- With the dielectric, the electric field becomes:
\[ E = \frac{E_0}{K} \]
- Thus, the electric field decreases.

6. **Effect on Voltage:**
- The voltage \( V \) across the capacitor is related to the electric field and the distance between the plates:
\[ V = E \cdot d \]
- Therefore, the new voltage with the dielectric is:
\[ V = \frac{V_0}{K} \]
- This indicates that the voltage decreases.

7. **Effect on Energy Stored:**
- The energy \( U \) stored in the capacitor is given by:
\[ U = \frac{1}{2} C V^2 \]
- Substituting the new capacitance and voltage:
\[ U = \frac{1}{2} (K C_0) \left(\frac{V_0}{K}\right)^2 = \frac{1}{2} K C_0 \frac{V_0^2}{K^2} = \frac{1}{2} \frac{C_0 V_0^2}{K} \]
- This shows that the energy stored decreases by a factor of \( K \).

### Summary of Effects:
1. **Capacitance increases**: \( C = K \cdot C_0 \)
2. **Electric field decreases**: \( E = \frac{E_0}{K} \)
3. **Voltage decreases**: \( V = \frac{V_0}{K} \)
4. **Energy stored decreases**: \( U = \frac{U_0}{K} \)